Answers for MATH0016
1.
Question 1
(a) The general form of the linear second order PDE for a function
φ
(
x, y
) is
Aφ
xx
+
Bφ
xy
+
Cφ
yy
=
D
(
x, y
)
(i)
A
= 1
,
B
= 1
,
C
=

2
Therefore
B
2

4
AC
= 9
>
0 and the PDE is hyperbolic.
(ii)
A
= 1
,
B
= 0
,
C
= 1
Therefore
B
2

4
AC
=

4
<
0 and the PDE is elleptic.
(iii)
A
= 1
,
B
=

2
,
C
= 1
Therefore
B
2

4
AC
= 0 and the PDE is parabolic
(b) Using the chain rule we have
∂
∂s
=
∂x
∂s
∂
∂x
+
∂y
∂s
∂
∂y
=
∂
∂x

β
∂
∂y
∂
∂t
=
∂x
∂t
∂
∂x
+
∂y
∂t
∂
∂y
=
∂
∂x

α
∂
∂y
so
∂
2
∂s∂t
= (
∂
∂x

β
∂
∂y
)(
∂
∂x

α
∂
∂y
) =
∂
2
∂x
2

(
α
+
β
)
∂
2
∂x∂y
+
αβ
∂
2
∂y
2
=
∂
2
∂x
2
+
∂
2
∂x∂y

2
∂
2
∂y
2
Therefore
α
+
β
=

1
and
αβ
=

2
It follows that
α
= 1
and
β
=

2
The change of variables is then
x
=
s
+
t,
y
= 2
s

t
and we have the simplified equation
∂
2
φ
∂s∂t
=
x
(c) Inverting the change of variable in (b) gives
s
=
x
+
y
3
1